A projectile is projected from ground with initial velocity `vecu=u_(0)hati+v_(0)hatj`. If acceleration due to gragvity (g) is along the negative y-direction then find maximum displacement in x-direction.

Options

A. `(u_(0)^(2))/(2g)`
B. `(2u_(0)v_(0))/(g)`
C. `(v_(0)^(2))/(2g)`
D. `(4u_(0)v_(0))/(g)`

Explanation

To solve the problem of finding the maximum displacement in the x-direction for a projectile projected from the ground with an initial velocity \( \vec{u} = u_0 \hat{i} + v_0 \hat{j} \), we can follow these steps:

### Step 1: Understand the motion components
The initial velocity has two components:
- Horizontal component: \( u_x = u_0 \)
- Vertical component: \( u_y = v_0 \)

### Step 2: Determine the time of flight
The time of flight \( T \) can be determined by analyzing the vertical motion. The vertical motion is influenced by gravity, which acts downwards. The equation of motion in the vertical direction is given by:

\[ y = u_y t - \frac{1}{2} g t^2 \]

At the maximum height, the vertical displacement \( y \) will be 0 (since it returns to the ground). Therefore, we set the equation to zero:

\[ 0 = v_0 t - \frac{1}{2} g t^2 \]

Factoring out \( t \):

\[ t (v_0 - \frac{1}{2} g t) = 0 \]

This gives us two solutions:
1. \( t = 0 \) (at the time of projection)
2. \( t = \frac{2v_0}{g} \) (time of flight)

### Step 3: Calculate the maximum displacement in the x-direction
The horizontal displacement \( x \) can be calculated using the horizontal component of the velocity and the time of flight:

\[ x = u_x \cdot T \]

Substituting the values we have:

\[ x = u_0 \cdot \frac{2v_0}{g} \]

### Final Expression
Thus, the maximum displacement in the x-direction is:

\[ \text{Maximum Displacement in x-direction} = \frac{2u_0 v_0}{g} \]